Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2011 / Paper 2 / 14 / a / iii / Solution

For $\lambda =5$, the eigenvalue equation reduces to a one-dimensional eigenspace spanned by $(1,2,-1)$. For $\lambda =-3$, the rows of $A+3I$ are multiples of $(1,2,-3)$, so its eigenspace is the plane $x+2y-3z=0$. One possible set of three independent eigenvectors isEvery eigenvector for $-3$ is a nonzero linear combination of the last two. Their two-dimensional eigenspace shows that the repeated root is not defective: the matrix is diagonalizable.